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Question: A car is moving with a speed of \[40\,{\text{km/hr}}\]. If the car engine generates 7 kilowatt power...

A car is moving with a speed of 40km/hr40\,{\text{km/hr}}. If the car engine generates 7 kilowatt power, then the resistance in the path of motion of the car will be
1. 360newton360\,{\text{newton}}
2. 630newton630\,{\text{newton}}
3. Zero
4. 280newton280\,{\text{newton}}

Explanation

Solution

Use the relation between the power generated, force and speed of the object. Relate the resistance in the path of motion of the car with the force against which the car engine have to generate the power.

Formula used: The relation between the power generated, force and speed of an object is
P=FvP = Fv …… (1)

Here, PP is the power generated, FF is the force on the object and vv is the speed of the object.

Complete step by step answer:
The car is moving with a speed vv of 40km/hr40\,{\text{km/hr}}.
v=40km/hrv = 40\,{\text{km/hr}}

Convert the speed vv of the car in the SI system of units.
v=(40kmhr)(103m1km)(1hr3600s)v = \left( {40\,\dfrac{{{\text{km}}}}{{{\text{hr}}}}} \right)\left( {\dfrac{{{{10}^3}\,{\text{m}}}}{{1\,{\text{km}}}}} \right)\left( {\dfrac{{1\,{\text{hr}}}}{{3600\,{\text{s}}}}} \right)
v=11.11m/s\Rightarrow v = 11.11\,{\text{m/s}}

Hence, the speed of the car is 11.11m/s11.11\,{\text{m/s}}.

The power PP generated by the engine of the car is 7kW7\,{\text{kW}}.
P=7kWP = 7\,{\text{kW}}

Convert the unit of the power PP generated by the engine in the SI system of units.
P=(7kW)(1031k)P = \left( {7\,{\text{kW}}} \right)\left( {\dfrac{{{{10}^3}}}{{1\,{\text{k}}}}} \right)
P=7×103W\Rightarrow P = 7 \times {10^3}\,{\text{W}}

Hence, the power generated by the car engine is 7×103W7 \times {10^3}\,{\text{W}}.

The power is generated by the car engine against the force to maintain the constant speed.

This force against which the car engine generates the power to maintain the speed constant is the resistance RR in the path of motion of the car.

Determine the resistance in the path of motion of the car.

Rewrite equation (1) for the resistance in the path of motion of the car.
P=RvP = Rv

Rearrange the above equation for the resistance RR.
R=PvR = \dfrac{P}{v}

Substitute 7×103W7 \times {10^3}\,{\text{W}} for PP and 11.11m/s11.11\,{\text{m/s}} for vv in the above equation.
R=7×103W11.11m/sR = \dfrac{{7 \times {{10}^3}\,{\text{W}}}}{{11.11\,{\text{m/s}}}}
R=630N\Rightarrow R = 630\,{\text{N}}

Therefore, the resistance in the path of motion of the car is 630N630\,{\text{N}}.

Hence, the correct option is 2.

Note:
Convert the units of power generated by the engine and the speed of the car in the SI system of units as the ultimate answer should be in newton.